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Python Library Reference |  |
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Python Library Reference | |
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The SequenceMatcher class has this constructor:
| [isjunk[, a[, b]]]) |
None (the default) or
a one-argument function that takes a sequence element and returns
true if and only if the element is ``junk'' and should be ignored.
Passing None for b is equivalent to passing
lambda x: 0; in other words, no elements are ignored. For
example, pass:
lambda x: x in " \t"
if you're comparing lines as sequences of characters, and don't want to synch up on blanks or hard tabs.
The optional arguments a and b are sequences to be compared; both default to empty strings. The elements of both sequences must be hashable.
SequenceMatcher objects have the following methods:
| a, b) |
SequenceMatcher computes and caches detailed information about the second sequence, so if you want to compare one sequence against many sequences, use set_seq2() to set the commonly used sequence once and call set_seq1() repeatedly, once for each of the other sequences.
| a) |
| b) |
| alo, ahi, blo, bhi) |
a[alo:ahi]
and b[blo:bhi].
If isjunk was omitted or None,
get_longest_match() returns (i, j,
k) such that a[i:i+k] is equal
to b[j:j+k], where
alo <= i <= i+k <= ahi and
blo <= j <= j+k <= bhi.
For all (i', j', k') meeting those
conditions, the additional conditions
k >= k',
i <= i',
and if i == i', j <= j'
are also met.
In other words, of all maximal matching blocks, return one that
starts earliest in a, and of all those maximal matching blocks
that start earliest in a, return the one that starts earliest
in b.
>>> s = SequenceMatcher(None, " abcd", "abcd abcd") >>> s.find_longest_match(0, 5, 0, 9) (0, 4, 5)
If isjunk was provided, first the longest matching block is determined as above, but with the additional restriction that no junk element appears in the block. Then that block is extended as far as possible by matching (only) junk elements on both sides. So the resulting block never matches on junk except as identical junk happens to be adjacent to an interesting match.
Here's the same example as before, but considering blanks to be junk.
That prevents ' abcd' from matching the ' abcd' at the
tail end of the second sequence directly. Instead only the
'abcd' can match, and matches the leftmost 'abcd' in
the second sequence:
>>> s = SequenceMatcher(lambda x: x==" ", " abcd", "abcd abcd") >>> s.find_longest_match(0, 5, 0, 9) (1, 0, 4)
If no blocks match, this returns (alo, blo, 0).
| ) |
(i, j, n), and
means that a[i:i+n] ==
b[j:j+n]. The triples are monotonically
increasing in i and j.
The last triple is a dummy, and has the value (len(a),
len(b), 0). It is the only triple with n == 0.
>>> s = SequenceMatcher(None, "abxcd", "abcd") >>> s.get_matching_blocks() [(0, 0, 2), (3, 2, 2), (5, 4, 0)]
| ) |
(tag, i1, i2,
j1, j2). The first tuple has i1 ==
j1 == 0, and remaining tuples have i1 equal to the
i2 from the preceeding tuple, and, likewise, j1 equal to
the previous j2.
The tag values are strings, with these meanings:
| Value | Meaning |
|---|---|
'replace' |
a[i1:i2] should be
replaced by b[j1:j2]. |
'delete' |
a[i1:i2] should be
deleted. Note that j1 == j2 in
this case. |
'insert' |
b[j1:j2] should be
inserted at a[i1:i1].
Note that i1 == i2 in this
case. |
'equal' |
a[i1:i2] ==
b[j1:j2] (the sub-sequences are
equal). |
For example:
>>> a = "qabxcd"
>>> b = "abycdf"
>>> s = SequenceMatcher(None, a, b)
>>> for tag, i1, i2, j1, j2 in s.get_opcodes():
... print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" %
... (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2]))
delete a[0:1] (q) b[0:0] ()
equal a[1:3] (ab) b[0:2] (ab)
replace a[3:4] (x) b[2:3] (y)
equal a[4:6] (cd) b[3:5] (cd)
insert a[6:6] () b[5:6] (f)
| [n]) |
Starting with the groups returned by get_opcodes(), this method splits out smaller change clusters and eliminates intervening ranges which have no changes.
The groups are returned in the same format as get_opcodes(). New in version 2.3.
| ) |
Where T is the total number of elements in both sequences, and M is
the number of matches, this is 2.0*M / T. Note that this is
1.0 if the sequences are identical, and 0.0 if they
have nothing in common.
This is expensive to compute if get_matching_blocks() or get_opcodes() hasn't already been called, in which case you may want to try quick_ratio() or real_quick_ratio() first to get an upper bound.
| ) |
This isn't defined beyond that it is an upper bound on ratio(), and is faster to compute.
| ) |
This isn't defined beyond that it is an upper bound on ratio(), and is faster to compute than either ratio() or quick_ratio().
The three methods that return the ratio of matching to total characters can give different results due to differing levels of approximation, although quick_ratio() and real_quick_ratio() are always at least as large as ratio():
>>> s = SequenceMatcher(None, "abcd", "bcde") >>> s.ratio() 0.75 >>> s.quick_ratio() 0.75 >>> s.real_quick_ratio() 1.0
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Python Library Reference | |
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